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3 years ago

Find an equation for the line below

Mathematics

1 answer:

dimulka [17.4K]3 years ago

5 0

Answer:

see the picture

Step-by-step explanation:

you can use geogebra to help you

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Solve for x. (x - 5)(x - 5) = 0

OverLord2011 [107]

Answer:

x=5

Step-by-step explanation:

(x - 5)(x - 5) = 0

x-5=0

x=5

6 0

3 years ago

N= 1•2•3•4•.......•49•50+23 <br> N:24=?

nadya68 [22]

Answer:

24347567557

Step-by-step explanation:

3 0

3 years ago

3 parentheses 2 minus 2 parentheses 3 parentheses 20 minus 2 parentheses

Monica [59]

Answer:

3(2-2) * 3(20-2)

3(0) * 3(18)

0 * 54 = 0

Step-by-step explanation:

3(2-2) * 3(20-2)

3(0) * 3(18)

0 * 54 = 0

5 0

3 years ago

Find the complete factored form of the

Andreas93 [3]

Answer:

$2a(-7ab^6+8)$

Step-by-step explanation:

2a is a common factor of the two terms:

$-14a^2b^6+16a=\boxed{2a(-7ab^6+8)}$

4 0

3 years ago

Calculus 3 chapter 16

o-na [289]

Evaluate $\vec F$ at $\vec r$ :

$\vec F(x,y,z) = x\,\vec\imath + y\,\vec\jmath + xy\,\vec k \\\\ \implies \vec F(\vec r(t)) = \vec F(\cos(t), \sin(t), t) = \cos(t)\,\vec\imath + \sin(t)\,\vec\jmath + \sin(t)\cos(t)\,\vec k$

Compute the line element $d\vec r$ :

$d\vec r = \dfrac{d\vec r}{dt} dt = \left(-\sin(t)\,\vec\imath+\cos(t)\,\vec\jmath+\vec k\bigg) \, dt$

Simplifying the integrand, we have

$\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt$

Then the line integral evaluates to

$\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}$

3 0

2 years ago

Find an equation for the line below​

Find an equation for the line below